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Supplementary note · 24 October 2026

Geometric arithmetic compatibility and recovery

The same forced discrepancy corrects every correspondingly labeled point. Valid ordered witnesses recover every finite integer history. The written arguments are supplied. They have not been reviewed as a proof, and this note does not amend Paper VII or frozen SRSD.

Start with the unit circle and the opening, then the declared bilateral extensions. The site’s selected outer readout is the average of the two equal golden segments, so G(a)=(aφ+aφ)/2=aφG(a)=(a\varphi+a\varphi)/2=a\varphi and C(a)=aπC(a)=a\pi. CC is the opposite semicircle, not the entire retained major arc. A seed chord is not part of the arc-and-feet curve. λ=φ/(φ+π)\lambda=\varphi/(\varphi+\pi) is the selected evolution factor.

Affine conjugacy

For a fixed positive calibration arefa_{\mathrm{ref}}, χ(a)=1+a/aref\chi(a)=1+a/a_{\mathrm{ref}} satisfies

χ(λa)=1+λ(χ(a)−1).\chi(\lambda a)=1+\lambda(\chi(a)-1).

This is a conjugacy on positive scales and their scalar image (1,∞)(1,\infty). It is not a claim about unlabeled set addition.

Pointwise discrepancy

Q=339949771344778Q=339949771344778, S=1015S=10^{15}, and q=Q/Sq=Q/S are distinct from λ\lambda. For a nonnegative integer dd, write d′=⌊qd⌋d'=\lfloor qd\rfloor and ρ=Qd−Sd′\rho=Qd-Sd', so 0≤ρ<S0\le\rho<S. For every fixed labeled template point pp,

q(dp)=d′p+ρSp.q(dp)=d'p+\frac{\rho}{S}p.

Divide Qd=Sd′+ρQd=Sd'+\rho by SS, then multiply by pp. The same scalar corrects every same-label point and every degree-one scale readout. That is pointwise equality, not a Minkowski sum of unlabeled sets.

Keep the ideal factor separate from the rational implementation. With Δ=λ−q>0\Delta=\lambda-q>0,

λ(dp)=d′p+(ρS+Δd)p.\lambda(dp)=d'p+\left(\frac{\rho}{S}+\Delta d\right)p.

The actual-factor discrepancy is ρ/S+(λ−q)d\rho/S+(\lambda-q)d. Floor agreement with λ\lambda is a certificate on 0≤d≤327670\le d\le 32767. One stated counterexample is d=42189065d=42189065, where the rational floor is 1434216214342162. The common-prefix statement on 0≤d0\le d below that counterexample is not an enumerated 42-million loop.

Inverse and ordered history

A valid pair (d′,ρ)(d',\rho) recovers its predecessor by

d=Sd′+ρQ.d=\frac{Sd'+\rho}{Q}.

Validity requires a nonnegative integer output, an integer remainder in [0,S)[0,S), and divisibility by QQ. That identity is the inverse on a valid image. It is not, by itself, a local certificate for every real trajectory. For every finite horizon NN,

QNd0=SNdN+∑k=0N−1ρkSkQN−1−k.Q^{N}d_{0}=S^{N}d_{N}+\sum_{k=0}^{N-1}\rho_{k}S^{k}Q^{N-1-k}.

Backward division recovers every intermediate state. With ek=ρk/Se_k=\rho_k/S,

λNd0−dN=∑k=0N−1λN−1−k(ek+Δdk).\lambda^{N}d_{0}-d_{N}=\sum_{k=0}^{N-1}\lambda^{N-1-k}(e_{k}+\Delta d_{k}).

The ideal and quantized trajectories remain distinct. The recovered initial state regenerates either one. Horizon 23 is a declared finite audit horizon, not a boundary of the written argument.

The same scalar discrepancy and remainder recovery extend, under a separately declared coupling, to the 600-cell signal network. The planar geometry does not select that graph or uniquely force its weights.

For output yy, the predecessors are b(y),…,b(y+1)−1b(y),\ldots,b(y+1)-1 with b(y)=⌈Sy/Q⌉b(y)=\lceil Sy/Q\rceil. Since 1/3<q<1/21/3<q<1/2, there are two or three. The rank w=d−b(y)∈{0,1,2}w=d-b(y)\in\{0,1,2\} is the conditional index. Two bits are the stated optimal worst-case fixed width given the output. That is not a claim about total compression.


Complete proofs and what was executed here

The complete written development is the restored v2 document: sixteen numbered theorems, the combined theorem, and the embedded Python and Java. Python is 14,819 bytes, 6d845f5646fcd9074d17b9eefca3c77640ea7e832aa70ca4f7b6237462306654. Java is 23,623 bytes, ff46f41d7246786adfda2e6bb0107e20e56f125c7319df176202c89216db2330. The v2 document’s own historical account is Python 3.13.5 / Java 21. The conversation reproduction used Python 3.12.14 and Java 17.0.20.

This archive ran both programs on 24 October 2026. Python 3.10.21, normal and python3 -O. OpenJDK 17.0.20, default and -ea, compiled with javac --release 17 -Xlint:all and no diagnostics. All four runs exited 0. Those stdout files are byte-identical, SHA-256 3c727ff3d73c936d26b97a5634d612a983beb296d8519893388e79e2bdea0996, and they match EXPECTED.txt. Stderr is empty.

Reported finite counts, not one undifferentiated proof total: 11 symbolic polynomial identities; 32,768 certified λ-floor states; 1 out-of-domain disagreement; 50,001 consecutive output fibers; maximum fiber cardinality 3; 2 optimal fixed-width witness bits; extinction horizon at most 10 on the audited seeds. No proof assistant. No peer review.

The hosted logs have recorded SHA-256 digests. Matching those digests checks the logs’ byte integrity. The logs document the archive’s reported executions. Rerunning the checker tests whether those results are reproducible. A matching hash does not itself verify that the recorded execution occurred or that the checker is mathematically correct.